Generalisation makes things better!

Let $\mathfrak{L}$ denote the given limit. Writing $\color{#3D99F6}{x^2+56x+1=(x-\alpha)(x-\beta)}$ and using $\small{\color{#3D99F6}{\lim_{x\rightarrow 0}\dfrac{1-\cos x}{x^2}=\dfrac{1}{2}}}$

$\large \mathfrak{L}= \lim_{x \to \beta} (\frac{1 - \cos (\color{#D61F06}{(x-\alpha)(x-\beta)})}{(\color{#D61F06}{(x-\alpha)^2(x - \beta)^2})}(x-\alpha)^2)$ $\Large =\dfrac{(\beta-\alpha)^2}{2}=\dfrac{(\beta+\alpha)^2}{2}-2\alpha\beta$ $\large =\dfrac{(-56)^2}{2}-2~~~(\small{\color{#20A900}{\because \alpha+\beta=-56, \alpha\beta=1})}$ $\huge =\boxed{\color{#007fff}{1566}}$

$\lim_{x \to \beta} \frac{1-\cos(x^2+56x+1)}{(x-\beta)^2}$

The above limit has immediate form $\frac{0}{0}$ hence we can apply the L'Hospital rule ,

Using the L'Hospital rule,

$\lim_{x \to \beta} \frac{\sin(x^2+56x+1)(2)(x+28)}{2(x-\beta)}$

Again applying the rule,

$\lim_{x \to \beta} 2(x+28)^2\cos(x^2+56x+1)+\sin(x^2+56x+1)$

$\lim_{x \to \beta} 2(x^2+56x+1+783)\cos(x^2+56x+1)+\sin(x^2+56x+1)$

$2(0+783)(1)+0$

$\boxed{1566}$

$L = \displaystyle \lim_{x\rightarrow \beta} \dfrac{1-\cos \left(x^{2}+56x+1\right)}{(x-\beta)^{2}} = \displaystyle \lim_{x\rightarrow \beta} \dfrac{2\sin^{2}\left(\dfrac{x^{2}+56x+1}{2}\right)}{(x-\beta)^{2}}$
Multiplying and dividing by $\dfrac{x^{2}+56x+1}{4}$
$L = \displaystyle \lim_{x\rightarrow \beta} \dfrac{2 \sin^{2} \left(\dfrac{x^{2}+56x+1}{2}\right)}{\left(\dfrac{x^{2}+56x+1}{2}\right)^{2}}\cdot \dfrac{2\left(x^{2}+56x+1\right)^{2}}{4(x-\beta)^{2}}$
Provided that both the limits are individually finite, we can split them as follows,
$L = \displaystyle \lim_{f(x)\rightarrow 0} \left(\dfrac{\sin f(x)}{f(x)}\right)^{2} \displaystyle \lim_{x\rightarrow \beta} \dfrac{2\left(x^{2}+56x+1\right)^{2}}{4(x-\beta)^{2}} = 1 \times\lim_{x\rightarrow \beta} \dfrac{\left(x^{2}+56x+1\right)^{2}}{2(x-\beta)^{2}}$
Substituting, $x^{2} + 56x + 1 = (x-\alpha)(x-\beta)$ ,
$L = \displaystyle \lim_{x\rightarrow \beta} \dfrac{(x-\alpha)^{2}(x-\beta)^{2}}{2(x-\beta)^{2}} = \displaystyle \lim_{x\rightarrow \beta} \dfrac{(x-\alpha)^{2}}{2} = \dfrac{(\beta-\alpha)^{2}}{2}$
For a given quadratic equation,
$ax^{2} + bx + c = 0$ with roots $\alpha , \beta$
$\beta - \alpha = \pm \dfrac{\sqrt{b^{2}-4ac}}{a}$
$\therefore L = \dfrac{\left(\dfrac{\pm \sqrt{56^{2}-4}}{1}\right)^{2}}{2} = \dfrac{56^{2}-4}{2} = \dfrac{3136-4}{2} = \dfrac{3132}{2} = 1566$